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Estadística y muestreo, 12ª.ed. (Segunda reimpresión) – CD Cap.6 Distribuciones de probabilidad Ciro Martínez Bencardino – Ecoe Ediciones Actualizado en diciembre de 2007 6 Distribuciones de probabilidad Distribución binomial – de Poisson – Hipergeométrica y normal EJERCICIOS RESUELTOS Se presenta el desarrollo de los 210 ejercicios que tiene este capítulo 1. Solución: ( 29 % 5 , 37 375 , 0 16 6 2 1 2 1 2 4 2 4 2 2 = = = = - = C P x 2 2 1 2 1 4 = = = = X q p n ( 29 % 5 , 37 2 = = x P (exactamente dos caras) 2. Solución: ( 29 1 3 4 3 3 2 1 2 1 = = C P x ( 29 % 25 25 , 0 16 4 16 1 4 2 1 2 1 ! 1 ! 3 ! 4 3 = = = = = = x P 3 2 1 2 1 4 = = = = X q p n ( 29 % 0 , 25 3 = = x P (exactamente 3 caras)

Capítulo CD 6 _(1 - 20_)

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  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    6 Distribuciones de probabilidad

    Distribucin binomial de Poisson Hipergeomtrica y normal

    EJERCICIOS RESUELTOS

    Se presenta el desarrollo de los 210 ejercicios que tiene este captulo

    1. Solucin:

    ( ) %5,37375,0166

    21

    21 2424

    22 ===

    =

    =CP x

    22121

    4

    =

    =

    =

    =

    Xqpn

    ( ) %5,372 ==xP

    (exactamente dos caras)

    2. Solucin:

    ( )13

    433 2

    121

    ==

    CP x

    ( ) %2525,0164

    1614

    21

    21

    !1!3!4

    3 ===

    =

    =

    =xP

    32121

    4

    =

    =

    =

    =

    X

    qpn

    ( ) %0,253 ==xP

    (exactamente 3 caras)

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    2

    3. Solucin:

    ( )

    =

    == 36

    25361

    !2!2!4

    65

    61 224

    22 CP x

    ( ) %57,111157,0296.1150

    296.1256

    296.125

    234

    2 ===

    =

    =

    =xP

    26561

    4

    =

    =

    =

    =

    X

    qpn

    ( ) %57,112 ==xP

    (exactamente dos cincos)

    4. Solucin:

    a) 8=n ( )ganarP 8,0= 2,0=q 2=X ( ) ?2 ==xP

    ( ) ( ) ( ) ( ) %1146,0001146,02,08,0 62822 ====xP ( ) %1146,02 ==xP

    b) 8=n ( )perderP 2,0= 8,0=q 2=X ( ) ?2 ==xP

    ( ) ( ) ( ) ( ) %36,292936,08,02,0 62822 ====xP ( ) %36,292 ==xP

    c) 8=n ( )perderP 2,0= 8,0=q

    8,7,6,5,4,3,2)2( ydosmnimox == ( ) ?2 =xP

    ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]

    ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]7181808021087654322

    8,02,08,02,01

    1

    +=

    +=++++++=

    x

    x

    P

    PPPPPPPPPP

    ( ) [ ] %67,494967,05033,013355,01678,012 ===+=xP ( ) %67,492 =xP

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    3

    d) 8=n ( )ganarP 8,0= 2,0=q 6,5,4,3,2,1,0 yX = ( ) ?6 =xP

    ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]088817876

    8765432106

    2,08,02,08,01

    1

    ++=

    +=++++++=

    x

    x

    P

    PPPPPPPPPP

    ( ) [ ] %67,494967,05033,011678,03355,016 ===+=xP ( ) %67,496 =xP

    e) 8=n ( )perderp 2,0= 8,0=q 6=X ( )6=xP

    ( ) ( ) ( ) ( ) %1147,0001147,08,02,0 26866 ====xP Observemos que decir: seis pierdan es lo mismo que dos ganen

    8=n ( )ganarp 8,0= 2,0=q 2=X ( )2=xP

    ( ) ( ) ( ) ( ) %1147,0001147,02,08,0 62822 ====xP ( ) %1147,02 ==xP

    5. Solucin:

    xnxnx qpCP

    = 5,021

    ==p 5,021

    ==q 6=n

    a) ( )

    =

    == 4

    1161

    !4!2!6

    21

    21 246

    44 CP x

    ( ) %44,232344,06415

    64115

    641

    256

    4 ===

    =

    =

    =xP ( ) %44,234 ==xP

    (exactamente 4 caras)

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    4

    b) Como mximo 4 caras

    ( )24

    64

    3363

    4262

    5161

    60604 2

    121

    21

    21

    21

    21

    21

    21

    21

    21

    +

    +

    +

    +

    = CCCCCP x

    ( ) ( ) ++++= 41

    161158

    1812016

    1411532

    121664

    1114xP

    ( ) %06,898906,06457

    6415

    6420

    6415

    646

    641

    4 ===++++=xP ( ) %06,894 =xP

    Tambin se puede resolver de la siguiente forma:

    ( )

    +

    =

    0666

    15654 2

    121

    21

    211 CCPx

    ( ) %06,898906,06457

    647

    6464

    646

    64114 ====

    +=xP ( ) %06,894 =xP

    (mximo 4 caras)

    6. Solucin:

    Aparicin de un cinco, la probabilidad es 1/6; Aparicin de un seis, la probabilidad es 1/6

    31

    62

    61

    61

    ==+=p 32

    31

    331 === pq

    a) ( ) %80,121280,0187.2280

    187.2835

    278

    811

    !3!4!7

    32

    31 34

    474 ===

    =

    =

    ==xP

    (cuatro xitos) ( ) %80,124 ==xP

    b) ( )34

    74

    6171

    70704 3

    231

    ..............32

    31

    32

    31

    +

    +

    = CCCP x

    ( ) ( ) ++++= 278

    8113581

    1627135243

    329121729

    64317187.2

    128114xP

    ( ) ==++++= 187.2088.2

    187.2280

    187.2560

    187.2672

    187.2448

    187.2128

    4xP

    %47,959547,0 == (mximo 4 xitos) ( ) %47,954 =xP

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    5

    Tambin puede resolverse as:

    ( )

    +

    +

    =

    071625

    4 32

    317

    732

    317

    632

    317

    51xP

    ( ) ( )

    +

    +

    = 1187.2113

    2729179

    424312114xP

    ( ) 0453,01187.2991187.2

    1187.214

    187.28414 ==

    ++=xP

    %47,959547,0 == ( ) %47,954 =xP

    7. Solucin:

    4=n 10,0=p 90,0=q

    a) ( ) ( ) ( ) ( ) ( ) %61,656561,06561,0119,01,0 40400 ===== CP x ( ) %61,650 ==xP

    b) ( ) ( ) ( ) ( ) ( ) %16,292916,0729,01,049,01,0 31411 ===== CP x ( ) %16,291 ==xP

    c) ( ) ( ) ( ) ( ) ( ) %86,40486,081,001,069,01,0 22422 ===== CP x ( ) %86,42 ==xP

    d) ( ) ( ) ( ) ( ) ( ) ( ) ( )2242314140402 9,01,09,01,09,01,0 CCCP x ++=

    ( ) %63,999963,00486,02916,06561,02 ==++=xP ( ) %36,992 =xP (no ms de dos defectuosos)

    8. Solucin:

    a) 40,0=p 60,0=q 5=n 2=X

    ( ) ( ) ( ) ( ) ( ) %56,343456,0216,016,0106,04,0 32522 ===== CP x ( ) %56,342 ==xP

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

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    b) ( ) ( ) ( ) ( ) ( )415150501 6,04,06,04,0 CCP x +=

    ( ) ( ) ( ) ( ) ( )1296,04,0!4!1!507776,01!5!0

    !51 +=xP

    ( ) ( ) ( ) ( ) ( ) 2592,007776,01296,04,0507776,0111 +=+=xP

    %69,333369,0 == (menos de 2 golpes) ( ) %69,331 =xP

    9. Solucin:

    8=n 5,0=p 5,0=q ,5,4,3,2,1,0=X

    ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )53836282718180805 5,05,05,05,05,05,05,05,0 CCCCP x +++=

    ( ) ( ) ( ) ( ) %54,8585543,05,05,05,05,0 35854484 ==++ CC ( ) %54,855 =xP

    Es posible resolverlos de la siguiente forma:

    ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]0888178726865 5,05,05,05,05,05,01 CCCP x ++=

    ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]100396,015,000781,0825,0015625,02815 ++=xP

    ( ) [ ] %54,8585543,014457,0100396,003124,010937,015 ===++=xP ( ) %54,855 =xP (menos de 6 caras)

    10. Solucin:

    05,0=p 95,0=q 6=n ,2,1,0=X

    ( ) ( ) ( ) ( ) ( ) ( ) ( )4262516160602 95,005,095,005,095,005,0 CCCP x ++=

    ( ) ( ) ( ) ( ) ( ) ( ) ( )814506,00025,015773780,005,06735091,0112 ++=xP

    ( ) %78,99997768,0030543,0232134,0735091,02 ==++=xP ( ) %78,992 =xP

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    7

    11. Solucin:

    10,0=p 90,0=q 5=n 0=X

    a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) %05,595905,05905,0111,09,09,01,0 055550500 ====== CCP x ( ) %05,590 ==xP

    b) ( ) ( ) ( ) ( ) ( ) ( ) ( )0555145423533 9,01,09,01,09,01,0 CCCP x ++=

    00856,000001,000045,000810,0 =++= ( ) %856,03 =xP

    c) ( ) ( ) ( ) %81,000810,09,01,0 23533 ==== CP x ( ) %81,03 ==xP (exactamente 3 mueran)

    12. Solucin:

    2,0=p 8,0=q 4=n

    a) ( ) ( ) ( ) ( ) ( ) %96,404096,0512,02,048,02,0 31411 ===== CP x ( ) %96,401 ==xP

    b) ( ) ( ) ( ) ( ) ( ) %96,404096,04096,0118,02,0 40400 ===== CP x ( ) %96,400 ==xP

    c) ( ) ( ) ( ) ( ) ( ) ( ) ( )2242314140402 8,02,08,02,08,02,0 CCCP x ++=

    ( ) %28,979728,01536,04096,04096,02 ==++=xP ( ) %28,972 =xP (no ms de dos cerrojos sean defectuosos)

    13. Solucin:

    4,0=p 6,0=q 5=n

    a) Que ninguno se gradu:

    ( ) ( ) ( ) %78,70778,06,04,0 50500 ==== CP x ( ) %78,70 ==xP

    b) Que se gradu uno:

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    8

    ( ) ( ) ( ) %92,252592,06,04,0 41511 ==== CP x ( ) %92,251 ==xP

    c) Que se grade al menos uno:

    ( ) ( ) ( ) %22,929222,00778,016,04,01 50501 ==== CP x ( ) %22,991 =xP

    14. Solucin:

    61=p 65=q 5=n

    a) ( ) %19,404019,0776.7125.3

    296.1625

    615

    65

    61 415

    11 ===

    =

    =

    =CP x ( ) %19,401 ==xP

    b) ( ) %08,161608,0776.7250.1

    216125

    361106

    561 325

    22 ===

    =

    ==

    CP x ( ) %08,162 ==xP

    c) ( ) %21,30321,0776.7250

    3625

    2161106

    561 235

    33 ===

    =

    ==

    CP x ( ) %21,33 ==xP

    d) ( ) %32,00032,0776.725

    65

    296.1156

    561 145

    44 ===

    =

    ==

    CP x ( ) %32,04 ==xP

    e) ( ) ( ) %19,404019,0776.7 125.311656150

    500 ==

    =

    ==

    CP x (ninguna vez) ( ) %19,400 ==xP

    15. Solucin:

    10,0=p 90,0=q 4=n

    a) ( ) ( ) ( ) %61,656561,09,01,0 40400 ==== CP x ( ) %61,650 ==xP

    b) ( ) ( ) ( ) %39,343439,09,01,01 40401 === CP x ( ) %39,341 =xP

    c) ( ) ( ) ( ) ( ) ( )314140401 9,01,09,01,0 CCP x +=

    %77,949477,02916,06561,0 ==+= ( ) %77,941 =xP 16. Solucin:

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    9

    2,0=p 8,0=q 10=n

    a) ( ) ( ) ( ) %2,303020,08,02,0 821022 ==== CP x ( ) %2,302 ==xP

    b) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]82102911011001003 8,02,08,02,08,02,01 CCCP x ++=

    ( ) [ ] %22,323222,06778,013020,02684,01074,013 ===++=xP ( ) %22,323 =xP

    c) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0101010191092810837107461066 8,02,08,02,08,02,08,02,08,02,0 CCCCCP x ++++=

    0063,00000,00000,00008,00055,0 =+++= ( ) %63,06 =xP (Se us la tabla para el clculo)

    d) ( ) ( ) ( ) %74,101074,08,02,0 1001000 ==== CP x ( ) %74,100 ==xP

    17. Solucin:

    5,0=p 5,0=q 10=n 0,1,2,3=X

    ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1001009110182102731033 5,05,05,05,05,05,05,05,0 CCCCP x +++=

    %19,171719,00010,00098,00439,01172,0 ==+++= ( ) %19,173 =xP

    npE = ( ) 100181719,0100 depersonasE =

    18. Solucin:

    5,0=p 5,0=q 10=n 10 9 ,8 ,7 yX =

    ( ) ( ) ( ) ( ) ( ) ( ) ( ) 01010101910928108371077 )5,0()5,0(5,05,05,05,05,05,0 CCCCP x +++=

    ( ) %19,171719,100010,00098,00439,01172,07 ==+++=xP ( ) %19,177 =xP

    19. Solucin:

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    10

    15=n 10,0=p 90,0=q

    a) ( ) ( ) ( ) %05,10105,09,01,0 1051555 ==== CP x ( ) %05,15 ==xP

    b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) +++= 31215124111511510151010 9,01,09,01,09,01,0 CCCP x

    ( ) ( ) ( ) ( ) ( ) ( ) 0000,09,01,09,01,09,01,0 015151511415142131513 =++ CCC ( ) 010 =xP

    (Como se trabaja con cuatro decimales, aproximamos a cero) (Se utiliz la tabla)

    A partir de x > 8 la probabilidad obtenida es demasiado pequea, casi cero.

    c) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ +++= 1321521411511501505 9,01,09,01,09,01,01 CCCP x

    ( ) ( ) ( ) ( ) ]114154123153 9,01,09,01,0 CC +

    Utilizando la tabla se tiene:

    ( ) [ ]9873,00428,01285,02669.03432,02059,015 =++++=xP ( ) %27,15 =xP

    ( ) %27,10127,09873,015 ===xP

    20. Solucin:

    20=n 25,0=p 75,0=q

    a) ( ) ( ) ( ) 0...............0000,075,025,0 515201515 ==== CP x (ver tabla) ( ) 015 ==xP

    b) ( ) ( ) ( ) ( ) ( ) ( ) ( )1642041912012002004 75,025,0...........75,025,075,025,0 CCCP x ++=

    %48,414148,01897,01339,00669,00211,00032,0 ==++++= ( ) %48,414 =xP

    c) ( ) ( ) ( ) ( ) ( ) ( ) ( )02020201192091282088 75,025,0...........75,025,075,025,0 CCCP x ++=

    Es ms fcil resolverlo de la siguiente forma:

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    11

    ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]1372071912012002008 75,025,0............75,025,075,025,01 CCCP x ++=

    [ ] =+++++++= 1124,01686,02023,01897,01339,00669,00211,00032,01

    %19,108981,01 == (por lo menos 8 defectuosas) ( ) %19,108 =xP

    21. Solucin:

    5,0=p 5,0=q 4=n

    a) ( ) ( ) ( ) 9375,00625,015,05,01 40401 === CP x ( ) %75,931 =xP

    ( ) 875.19375,0000.2 ==E familias

    b) ( ) ( ) ( ) 3750,05,05,0 22422 === CP x ( ) %50,372 ==xP

    ( ) familiasE 7503750,0000.2 ==

    c) ( ) ( ) ( ) 0625,05,05,0 40400 === CP x ( ) %25,60 ==xP

    ( ) familiasE 1250625,0000.2 ==

    (Se utilizaron las tablas)

    22. Solucin:

    ( ) ( ) ( ) ( ) ( ) ( ) ( )1321521411511501502 95,005,095,005,095,005,0 CCCP x ++=

    9639,01348,03658,04633,0 =++= = 96,39% ( ) %39,962 =xP

    (Se utiliz la tabla)

    23. Solucin:

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    12

    40,0=p 20=n

    ( ) ( ) ( ) ( ) ( ) ( ) ( )02020208122012911201111 6,04,0........6,040,06,04,0 CCCP x +++=

    Utilizando la tabla se tendr que:

    ( ) =+++++++++= 00000003,00013,00049,00146,00355,00710,011xP

    %76,121276,0 == (mitad ms uno) ( ) %76,1211 =xP

    (Se utiliz la tabla para el clculo)

    24. Solucin:

    20,0=p 80,0=q 18=n 8=X

    ( ) ( ) ( ) %20,10120,080,020,0 1081888 ==== CP x ( ) %20,18 ==xP

    25. Solucin:

    ( ) ( ) ( ) ( ) ( ) ( ) ( )37107461065510575 5,05,05,05,05,05,0 CCCP x ++=

    ( ) %84,565684,01172,02051,02461,075 ==++= xP ( ) %84,5675 = xP

    26. Solucin:

    5=n ( )3xP 5,4,3=X 5,0=p 5,0=q

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    13

    ( ) ( ) ( ) ( )5433 === ++= xxxx PPPP

    ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )055514542353 5,05,05,05,05,05,0 ++=

    %505000,003125,015625,03125,0 ==++= ( ) %503 =xP

    27. Solucin:

    cariescon 90,0109

    = %1010,0cariessin == 5=n

    a) Cuatro tengan caries 5=n 90,0=p 4=X

    ( ) ( ) ( ) ( ) %81,3232805,01,09,0 14544 ====xP ( ) %81,324 ==xP

    b) Por lo menos dos tengan caries 90,0=p 5,4,3,2=X

    ( ) ( ) ( ) ( ) ( )54322 ==== +++= xxxxx PPPPP

    ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )[ ]41515050

    10

    1,09,01,09,01

    1

    +=

    +=== xx PP

    [ ] %95,999995,000045,000001,01 =+= ( ) %95,992 =xP

    c) Por lo menos 2 no tengan caries: 10,0=p 5,4,3,2=X

    ( ) ( ) ( ) ( ) ( )54322 ==== +++= xxxxx PPPPP

    ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )[ ]41515050

    10

    9,01,09,01,01

    1

    +=

    +=== xx PP

    [ ] %15,89185,0132805,059049,01 ==+= ( ) %15,82 =xP

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    14

    d) Por lo menos una tenga caries 90,0=p 5,4,3,2,1=X

    ( ) ( )01 1 = = xx PP

    ( ) ( ) ( ) %10099999,000001,011,09,01 5050 ==== ( ) %1001 =xP

    28. Solucin:

    20% pierden el 1 ao pierden lo no 80% 6=n

    a) :aprueben 2 Mximo 210 , , X = 800,p =

    ( ) ( ) ( ) ( )2102 === ++= xxxx PPPP

    ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )4262516160602 2,08,02,08,02,08,0 ++=xP

    %70,101696,001536,0001536,0000064,0 ==++= ( ) %70,12 =xP

    b) Todos aprueben: 800, p = 6=X

    ( ) ( ) ( ) ( ) %21,262621,02,08,0 06666 ====xP ( ) %21,266 ==xP

    c) Ninguno apruebe 800, p = 0=X

    ( ) ( ) ( ) ( ) %0064,0000064,02,08,0 60600 ====xP ( ) %0064,00 ==xP

    29. Solucin:

    0,7000068004

    =

    .

    .

    Transporte pblico 30% 0,30 = otro servicio

    a) No ms de 2 utilicen transporte pblico 70,p = 2,1,0=X 8=n

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

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    ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )6282718180802 3,07,03,07,03,07,0 ++=xP

    %13,101129,001000,00012247,00000656,0 ==+== ( ) %13,12 =xP

    b) Por lo menos 3 no lo utilicen 30,0=p 8,7,6,5,4=X

    ( ) ( ) ( ) ( ) ( ) ( ) ( )8765433 ====== +++++= xxxxxxx PPPPPPP

    ( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]628271818080

    210

    7,03,07,03,07,03,01

    1

    ++=

    ++==== xxx PPP

    [ ] %82,,444482,02965,01977,00576,01 ==++= ( ) %82,443 =xP

    c) Exactamente 2 no lo utilicen 30,0=p 2=X

    ( ) ( ) ( ) ( ) %65,292965,07,03,0 62822 ====xP ( ) %65,292 ==xP

    d) Exactamente 2 lo utilicen 70,0=p 2=X

    ( ) ( ) ( ) ( ) %10100,03,07,0 62822 ====xP ( ) %12 ==xP

    30. Solucin:

    60% = 0,60 asisten 0,40 = 40% no asisten n = 8

    a) asistan 7 menos loPor 6,0=p 8,7=X

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    16

    ( ) ( ) ( )877 == += xxx PPP

    ( ) ( ) ( ) ( ) ( ) ( )08881787 4,06,04,06,0 +=

    %64,101064,00168,00896,0 ==+= ( ) %64,107 =xP

    b) Por lo menos 2 no asistan 8=n 40,0=p 8,7,6,5,4,3,2=X

    ( ) ( ) ( ) ( )8322 .................... === +++= xxxx PPPP

    ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )[ ]71818080

    10

    6,04,06,04,01

    1

    +=

    +=== xx PP

    [ ] %36,898936,01064,010896,00168,01 ===+= ( ) %36,892 =xP

    31. Solucin:

    gafasusan 4,02000800

    = gafasusan no0,6 = 5n =

    a) gafasusan 2 menos loPor 40,0=p 5,4,3,2=X

    ( ) ( ) ( ) ( ) ( )54322 ==== +++= xxxxx PPPPP

    ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )[ ]41515050

    10

    6,04,06,04,01

    1

    +=

    +=== xx PP

    [ ] %3,666630,03370,012592,00778,01 ===+= ( ) %30,662 =xP

    b) gafasusan no 2 menos loPor 60,0=p 5,4,3,2=X

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    17

    ( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )[ ][ ] %30,9191296,008704,010768,001024,01

    4,06,04,06,01

    1

    4151

    5050

    102

    ===+=

    +=

    +=== xxx PPP

    ( ) %30,912 =xP

    c) ( ) gafasusen no espera se alumnos,200.160,02000 === EnpE

    32. Solucin:

    repitentesson 33,031 = repitentes no0,67 = 4n =

    a) repitentessean dos de mas No 33,0=p 2,1,0=X

    ( ) ( ) ( ) ( )2102 === ++= xxxx PPPP

    ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )%18,898918,02933,03970,02015,0

    67,033,067,033,067,033,0 2242314

    1404

    0

    ==++=

    ++=

    ( ) %18,882 =xP 32y31con trabajamosS:Nota ( ) %89,882 =xP

    b) repitente sea no 1 menos Al 67,0=p 4,3,2,1=X

    ( ) ( ) ( ) ( ) ( )43211 ==== +++= xxxxx PPPPP

    32y31con os trabajamS :Nota ( ) %77,981 =xP

    ( ) ( )01 1 = = xx PP

    ( ) ( ) ( ) %81,989881,00119,0133,067,01 4040 ==== ( ) %81,981 =xP

    33. Solucin:

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    18

    16=n 6,0=p 16,15,14,13,12,11,10=X

    ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )313161341216125111611610161010 4,06,04,06,004,06,04,06,0 +++=xP

    ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) =+++ 016161611516152141614 4,06,04,06,04,06,0

    0150,00468,01014,01623,01983,0 ++++=

    %71,520003,00030,0 =++ ( ) %71,5210 =xP (diez o ms acontecimientos desfavorables)

    34. Solucin:

    accidentan se 25% accidentan se no 75%

    accidentan se 3 menos loPor 7=n

    ( ) ( ) ( ) ( ) ( ) ( )765433 ===== ++++= xxxxxx PPPPPP

    ( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]527261717070

    210

    75,025,075,025,075,025,01

    1

    ++=

    ++==== xxx PPP

    [ ] %35,242435,07565,013115,03115,01335,01 ===++= ( ) %35,243 =xP

    35. Solucin:

    3% son defectuosos 97% Buenos n = 7

    a) Por lo menos 3 sean buenos

    ( ) ( ) ( ) ( )7433 ....... === +++= xxxx PPPP

    ( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]527261717070

    210

    03,097,003,097,003,097,01

    1

    ++=

    ++==== xxx PPP

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    19

    [ ] %1001 a aproxima se0001 ==++= ( ) %1003 =xP

    b) Por lo menos 3 sean defectuosos

    ( ) ( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ][ ] %09,0%0009,09991,010162,01749,08080,01

    97,003,097,003,097,003,01

    1

    5272

    6171

    7070

    2103

    ===++=

    ++=

    ++==== xxxx PPPP

    ( ) %09,03 =xP

    36. Solucin:

    enferman01,0 ==p 5=n enferman no99,0 ==q

    a) enfermos2=X

    ( ) ( ) ( ) ( ) %097,000097,099,001,0 32522 ====xP ( ) %097,02 ==xP

    b) enfermo uno menos loPor 5432,1 , , , X =

    ( ) ( ) ( ) ( ) ( ) %9,4049,09510,0199,001,011 505001 ===== = xx PP ( ) %9,41 =xP

    c) Por lo menos 2 no enfermen 5432 , , , X =

    ( ) ( ) ( )[ ]( ) ( ) ( ) ( ) ( ) ( )[ ][ ] %1001 a aproxima se001

    01,099,001,099,01

    1

    4151

    5050

    102

    ==+=

    +=

    +=== xxx PPP

    ( ) %1002 =xP

    37. Solucin:

  • Estadstica y muestreo, 12.ed. (Segunda reimpresin) CD Cap.6 Distribuciones de probabilidad Ciro Martnez Bencardino Ecoe Ediciones Actualizado en diciembre de 2007

    20

    20% de mortalidad 80% de sobrevivir 5=n

    a) Ninguno sobreviva 0=X ( )mueran todos,5aequivale =x

    ( ) ( ) ( ) ( ) ( ) ( ) ( ) %032,000032,08,02,02,08,0 055550500 =====xP ( ) %032,00 ==xP

    b) Todos sobrevivan

    ( ) ( ) ( ) ( ) %77,323277,02,08,0 05555 ====xP ( ) %77,325 ==xP

    c) Al menos 1 sobrevivan 54321 , , , , X =

    ( ) ( ) ( ) ( ) ( ) %97,99%968,9900032,012,08,011 505001 ===== = xx PP ( ) %97,991 =xP

    d) Al menos 1 no sobrevivan 54321 , , , , X =

    ( ) ( ) ( ) ( ) ( ) %23,6767232,032768,018,02,011 505001 ===== = xx PP ( ) %23,671 =xP

    38. Solucin:

    scientfico 20%0,20255

    == cientfico no 80%2520

    = 4=n

    a) Por lo menos 1 sea cientfica 4,3,2,1=X

    ( ) ( ) ( ) ( ) ( ) %04,595904,04096,018,02,011 404001 ===== = xx PP ( ) %04,591 =xP